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Thermodynamics

# Calculate $\Delta H_{C-H}$ from the following data: $\Delta H^{\circ}_{combustion}$ of ethane and propane are $-372$ and $-530\: kcal/mol$ respectively, $\Delta H_f^{\circ}$ of $CO_2(g)$ and $H_2O(l)$ are $-94$ and $-68\: kcal/mol,\: \Delta H^{\circ}$ for $C(graphite) \rightarrow C(g)\: is\: 172\: kcal \: and\: \Delta H_{H-H} \: is\: 104 \: kcal.$

$\begin {array} {1 1} (A)\;-99\: kcal/mol & \quad (B)\;99\: kcal/mol \\ (C)\; 82\: kcal/mol & \quad (D)\;- 82\: kcal/mol \end {array}$

$C_2H_6 (g) + \large\frac{7}{2}$$O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (l); \Delta H_1 = -372 \: kcal C_3H_8 (g) + 5O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l); \Delta H_2 = -530 \: kcal H_2 (g) + \large\frac{1}{2}$$ O_2 (g) \rightarrow H_2O (l); \: \: \: \:\: \: \:\: \: \: \:\: \: \: \: \: \: \: \Delta H_3 = -68 \: kcal$
$C(s) + O_2 (g) \rightarrow CO_2 (g); \: \: \: \:\: \: \:\: \: \: \:\: \: \: \: \: \: \: \Delta H_4 = -94\: kcal$
$H_2 (g) \rightarrow 2H (g); \: \: \: \:\: \: \:\: \: \: \:\: \: \: \: \: \: \:\: \: \: \: \: \: \: \Delta H_5 = 104\: kcal$
$C (graphite) \rightarrow C (g); \: \: \: \:\: \: \:\: \: \: \:\: \: \: \: \: \: \: \Delta H_6 = 172\: kcal$
$CH_3 – CH_3 \rightarrow 2C (g) + 6H (g); \Delta H_7$
$CH_3CH_2CH_3 (g) \rightarrow 3C (g) + 8H (g); \Delta H_8$
Thus, $\Delta H_7 = 2 \times \Delta H_6 + 3 \times \Delta H_5 + \Delta H_1 - \Delta H_4 - 3 \times \Delta H_3 = 2 \times 172 + 3 \times 104 + (-372) – (-94)- 3(-68) = 676kcal$
So, $\Delta H_7 = \Delta H_{C-C} +6 \Delta H_{C-H} = 676\: kcal$
Similarly, $\Delta H_8 = 3\Delta H_6 + 4\Delta H_5 + \Delta H_2 – 3\Delta H_4 – 4\Delta H_3 = 3(172) + 4(104)–530–3(-94) – 4(68) = 956\: kcal$
So, $\Delta H_8 = 2 \Delta H_{C-C} + 8 \Delta H_{C-H} = 956\: kcal$
$2 \Delta H_7 - \Delta H_8 = 2 \Delta H_{C-C} - 2 \Delta H_{C-C} + 12 \Delta H_{C-H} - 8 \Delta H_{C-H}$
$2(676) – 956 = 4 \Delta H_{C-H}\: or,\: \Delta H_{C-H} = \large\frac{396}{4}$$= 99\: kcal/mol$
Ans : (B)